truss_design_art

truss_design_art - Truss Design and Convex Optimization...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Truss Design and Convex Optimization Robert M. Freund April 13, 2004 c ± 2004 Massachusetts Institute of Technology. 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
1 Outline Physical Laws of a Truss System The Truss Design Problem Second-Order Cone Optimization Truss Design and Second-Order Cone Optimization Some Computational Results Semi-Definite Optimization Truss Design and Semi-Definite Optimization Truss Design and Linear Optimization Extensions of the Truss Design Problem 2 Physical Laws of a Truss System A truss is a structure in d =2 or d = 3 dimensions, formed by n nodes and m bars joining these nodes. Figure 1 shows an example of a truss. 2 3 4 5 6 1 Figure 1: Example of a truss in d = 2 dimensions, with n = 6 nodes and m =13 bars . Examples of trusses include bridges, cranes, and the Eiffel Tower. 2
Background image of page 2
The data used to describe a truss is: a set of nodes (given in physical space) a set of bars joining pairs of nodes with associated data for each bar: the length L k of bar k the Young’s modulus E k of bar k the volume t k of bar k an external force vector F on the nodes Nodes can be static (fixed in place) or free (movable when the truss is stressed). The allowable movements of the nodes defines the degrees of freedom (dof) of the truss. We say that the truss has N degrees of freedom. Of course, N nd . Movements of nodes in the truss will be represented by a vector u of N displacements , where u ∈± . N The external force on the truss is given by a vector F . Displacements of the nodes in the truss cause internal forces of compres- sion and/or expansion to appear in the bars in the truss. Let f k denote the internal force of bar k . The vector f m is the vector of forces of the bars. Figure 2 shows an example of a truss problem. In the figure, there is a single external force applied to node 3 in the direction indicated. This will result in internal forces along the bars in the truss and will simultaneously cause small displacements in all of the nodes. In Figure 2, nodes 1 and 5 are fixed, and the other nodes are free. To simplify notation we will label the 13 different bars by the nodes that hey link to: in general the bar k will join nodes i and j . The bars will be: 12, 13, ... , 56. The internal force f k of bar k can be positive or negative: If f k 0, bar k has been expanded and its internal force counteracts the expansion with compression, see Figure 3. 3
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2 3 4 5 6 1 f f F 13 12 3 Figure 2: A truss problem. > 0 L k Figure 3: A bar under expansion. L k ∆ < 0 Figure 4: A bar under compression. 4
Background image of page 4
If f k 0, bar k has been compressed and its internal force counteracts the compression with expansion, see Figure 4. 2.1 Physical Laws 2.1.1 Force Balance Equations In a static truss the internal forces will balance the external forces in every degree of freedom. That is, the forces around each free node must balance.
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 47

truss_design_art - Truss Design and Convex Optimization...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online